You have found the following ages (in years) of 6 turtles. Those turtles were randomly selected from the 32 turtles at your local zoo: $ 20,\enspace 32,\enspace 22,\enspace 86,\enspace 68,\enspace 28$ Based on your sample, what is the average age of the turtles? What is the standard deviation? You may round your answers to the nearest tenth.
Explanation: Because we only have data for a small sample of the 32 turtles, we are only able to estimate the population mean and standard deviation by finding the sample mean $({\overline{x}})$ and sample standard deviation $({s})$ To find the sample mean , add up the values of all $6$ samples and divide by $6$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\overline{x}} = \dfrac{20 + 32 + 22 + 86 + 68 + 28}{{6}} = {42.7\text{ years old}} $ Find the squared deviations from the mean for each sample. Since we don't know the population mean, estimate the mean by using the sample mean we just calculated {515.29} + {114.49} + {428.49} + {1874.89} + {640.09} + {216.09}} {{6 - 1}} $ {s^2} = \dfrac{{3789.34}}{{5}} = {757.87\text{ years}^2} $ As you might guess from the notation, the sample standard deviation $({s})$ is found by taking the square root of the sample variance $({s^2})$ ${s} = \sqrt{{s^2}}$ $ {s} = \sqrt{{757.87\text{ years}^2}} = {27.5\text{ years}} $ We can estimate that the average turtle at the zoo is 42.7 years old. There is also a standard deviation of 27.5 years.